A store sells two types of tables: plain and deluxe. When a customer buys a table, there is an 80% chance that it will be a plain table. Assume that whether or not the type of table that one customer buys is independent of the types of the tables that any other customer buys.
a. Suppose that on each of the days Monday, Tuesday, and Wednesday five tables are sold. What is the probability that all the tables sold on Monday and Tuesday are plain and that on Wednesday at least one deluxe table is sold?
b. Suppose the number of tables sold each day is a Poisson random variable with parameter x = 5. What is the probability that exactly five tables will be sold on a given day?
Probability - Reiny, Wednesday, January 21, 2009 at 12:35pm
Prob(5plain tables) on Monday = .8^5
prob(5 plain) on Tuesday = .8^5
prob(5 plain on Monday AND 5 plain on Tuesday) = (.8^5)(.8^5) = .8^10 = .107
Prob(at least one deluxe) = 1 - Prob(all 5 plain)
= 1 - .8^5 = .672
b. don't know about Poisson random variables.
Probability - economyst, Wednesday, January 21, 2009 at 1:24pm
A poisson distribution follows the formula: (e^(-x) * x^k)/k!
where x is the expected mean and k is the number of "true" events. You are given that the mean number of sales of tables is 5 (x=5) and you are asked for the probability that exactly 5 tables are sold (k=5)
Plug in the formula:
e^(-5) = .066738
5^5 = 3125
5! = 120
poisson=.1755 = 17.55%